Applications of Vectors

Vectors are especially useful when a real-world quantity has both a magnitude and a direction.

They are used to describe:

• displacement
• velocity
• acceleration
• force
• momentum
• wind and water currents
• aircraft and ship motion

Vector mathematics allows us to combine these quantities and determine their overall effect.

Vectors in Navigation

Navigation involves both distance and direction, so vectors provide a natural way to represent movement.

Suppose a boat travels:

8 km east

and then:

6 km north

Using components:

D₁ = (8, 0) km

D₂ = (0, 6) km

The resultant displacement is:

R = D₁ + D₂

R = (8, 0) + (0, 6)

R = (8, 6) km

Its magnitude is:

|R| = √(8² + 6²)

|R| = √100

|R| = 10 km

The boat finishes 10 km from its starting position.

Finding the Direction

For:

R = (8, 6)

use:

tan θ = 6/8

θ = tan⁻¹(6/8)

θ ≈ 36.9°

Therefore, the boat's displacement is:

10 km at 36.9° north of east

Notice that the total distance travelled was:

8 + 6 = 14 km

but the displacement magnitude was only:

10 km

Multiple Navigation Vectors

Suppose a hiker travels:

• 5 km east
• 4 km north
• 2 km west
• 1 km south

Choose:

east = +x

north = +y

Horizontal displacement:

Rₓ = 5 − 2

Rₓ = 3 km

Vertical displacement:

Rᵧ = 4 − 1

Rᵧ = 3 km

Therefore:

R = (3, 3) km

Magnitude:

|R| = √(3² + 3²)

|R| = √18

|R| ≈ 4.24 km

Direction:

tan θ = 3/3

θ = 45°

Therefore, the hiker finishes approximately:

4.24 km at 45° north of east

from the starting position.

Bearings

Navigation often describes direction using bearings.

A bearing is normally measured:

• clockwise
• from north
• using three digits

Examples:

000° = north

090° = east

180° = south

270° = west

Therefore, a direction such as:

045°

means northeast, halfway between north and east.

Vectors can be resolved into components when a bearing is known.

Resolving a Navigation Vector

Suppose a boat travels 20 km on a bearing of 060°.

Because bearings are measured from north:

east component:

Dₓ = 20 sin 60°

Dₓ ≈ 17.3 km

north component:

Dᵧ = 20 cos 60°

Dᵧ = 10.0 km

Therefore:

D ≈ (17.3, 10.0) km

This is an important difference from angles measured from the horizontal axis.

For an angle measured from the positive x-axis, we normally use:

x = R cos θ

y = R sin θ

For a bearing measured from north, the sine and cosine positions are reversed.

Aircraft and Wind

An aircraft's motion relative to the ground depends on both:

• its velocity through the air
• the velocity of the wind

These velocities must be added as vectors.

Suppose an aircraft flies east at:

200 km/h

while a wind blows north at:

50 km/h

Aircraft velocity:

V₁ = (200, 0) km/h

Wind velocity:

V₂ = (0, 50) km/h

Ground velocity:

V = V₁ + V₂

V = (200, 50) km/h

Finding the Aircraft's Ground Speed

The magnitude of the ground velocity is:

|V| = √(200² + 50²)

|V| = √42500

|V| ≈ 206 km/h

The direction is:

tan θ = 50/200

θ = tan⁻¹(0.25)

θ ≈ 14.0°

Therefore, the aircraft's ground velocity is approximately:

206 km/h at 14.0° north of east

Even though the pilot points the aircraft directly east, the wind causes the aircraft to drift north.

Correcting for Wind

Pilots may need to point the aircraft slightly into the wind so that its actual path over the ground follows the desired direction.

For example, if the wind pushes an aircraft north but the desired ground path is east, the aircraft may need to point slightly south of east.

This is called a wind correction.

The same vector principle applies to:

• ships affected by ocean currents
• swimmers crossing rivers
• drones flying in wind
• moving objects on conveyor belts

Boats and Water Currents

Suppose a boat travels north through the water at:

6 m/s

while a river flows east at:

8 m/s

Boat velocity:

V₁ = (0, 6) m/s

River velocity:

V₂ = (8, 0) m/s

Resultant velocity:

V = (8, 6) m/s

Magnitude:

|V| = √(8² + 6²)

|V| = 10 m/s

The boat's actual motion relative to the shore is therefore 10 m/s diagonally downstream.

Crossing a River

Suppose a river is 120 m wide.

A swimmer can swim directly across the river at:

3 m/s

while the current moves downstream at:

2 m/s

The time required to cross depends on the across-river component, not the total resultant speed.

Time:

t = distance / speed

t = 120 / 3

t = 40 s

During those 40 seconds, the river moves the swimmer downstream:

d = vt

d = 2 × 40

d = 80 m

Therefore, the swimmer reaches the opposite bank 80 m downstream from the point directly opposite the starting position.

Vectors and Forces

Force is a vector because it has both:

• magnitude
• direction

If several forces act on an object, their combined effect is called the resultant force or net force.

For example, suppose:

• 15 N acts east
• 9 N acts west

Choose east as positive.

Net horizontal force:

Fₓ = 15 − 9

Fₓ = 6 N east

Because the forces act along the same line, their effects can be combined directly.

Forces in Two Dimensions

Suppose an object experiences:

• 12 N east
• 5 N west
• 10 N north
• 4 N south

Horizontal component:

Fₓ = 12 − 5

Fₓ = 7 N

Vertical component:

Fᵧ = 10 − 4

Fᵧ = 6 N

Therefore:

F = (7, 6) N

Magnitude:

|F| = √(7² + 6²)

|F| = √85

|F| ≈ 9.22 N

Direction:

tan θ = 6/7

θ ≈ 40.6°

Therefore, the resultant force is:

9.22 N at 40.6° north of east

Resolving Forces into Components

Forces do not always act horizontally or vertically.

Suppose a force of 50 N acts at 30° above the horizontal.

Horizontal component:

Fₓ = 50 cos 30°

Fₓ ≈ 43.3 N

Vertical component:

Fᵧ = 50 sin 30°

Fᵧ = 25.0 N

Therefore:

F ≈ (43.3, 25.0) N

Resolving forces into components allows angled forces to be combined with other forces.

Worked Force Problem

Three forces act on an object:

• 20 N east
• 12 N north
• 10 N at 30° south of west

For the third force:

horizontal component:

F₃ₓ = −10 cos 30°

F₃ₓ ≈ −8.66 N

vertical component:

F₃ᵧ = −10 sin 30°

F₃ᵧ = −5.00 N

Now add the components.

Horizontal:

Rₓ = 20 − 8.66

Rₓ = 11.34 N

Vertical:

Rᵧ = 12 − 5

Rᵧ = 7.00 N

Resultant:

R = (11.34, 7.00) N

Magnitude:

|R| = √(11.34² + 7²)

|R| ≈ 13.3 N

Direction:

tan θ = 7/11.34

θ ≈ 31.7°

Therefore, the resultant is approximately:

13.3 N at 31.7° north of east

Balanced Forces

If the vector sum of all forces is zero:

F_net = (0, 0)

then the forces are balanced.

For example:

• 10 N east
• 10 N west
• 6 N north
• 6 N south

The components are:

Fₓ = 10 − 10 = 0

Fᵧ = 6 − 6 = 0

Therefore:

F_net = (0, 0)

The object has zero acceleration.

This does not necessarily mean the object is stationary. It could be moving at constant velocity.

Vectors and Newton's Second Law

Newton's Second Law is fundamentally a vector equation:

F_net = ma

Acceleration occurs in the same direction as the net force.

Suppose:

F_net = (12, 16) N

and:

m = 4 kg

Then:

a = F_net / m

a = (12/4, 16/4)

a = (3, 4) m/s²

Magnitude:

|a| = √(3² + 4²)

|a| = 5 m/s²

The acceleration has the same direction as the resultant force.

Velocity as a Vector

Velocity is different from speed.

Speed tells us how fast something moves.

Velocity tells us how fast and in what direction.

For example:

20 m/s

is a speed.

20 m/s east

is a velocity.

Velocity can therefore be represented using vector components.

Combining Velocities

Suppose:

V₁ = (12, 5) m/s

and:

V₂ = (−3, 4) m/s

Then:

V = V₁ + V₂

V = (12 − 3, 5 + 4)

V = (9, 9) m/s

Magnitude:

|V| = √(9² + 9²)

|V| = √162

|V| ≈ 12.7 m/s

Direction:

45° above the positive x-axis

Relative Velocity

Vectors are also used to describe motion relative to another moving object.

Suppose:

Car A travels east at:

25 m/s

Car B travels east at:

18 m/s

From Car A's perspective, Car B appears to move backward at:

18 − 25 = −7 m/s

The relative velocity of B compared with A is:

V_BA = V_B − V_A

This concept is important in:

• road traffic
• aircraft navigation
• spacecraft
• ships
• sports

Vectors and Acceleration

Acceleration is also a vector.

An object accelerates whenever its velocity changes.

This can happen because:

• its speed increases
• its speed decreases
• its direction changes
• both speed and direction change

For example, an object moving around a circular path may maintain constant speed while continually changing velocity because its direction changes.

Therefore, it is accelerating.

Vectors and Momentum

Momentum is defined as:

p = mv

Because velocity is a vector, momentum is also a vector.

Suppose a 5 kg object moves east at 8 m/s.

Its momentum is:

p = 5 × 8

p = 40 kg·m/s east

If it moved west instead, its momentum would point west.

Momentum conservation therefore requires us to account for direction.

Collision Example

Suppose:

• a 2 kg cart moves east at 5 m/s
• a 3 kg cart moves west at 2 m/s

Choose east as positive.

Cart 1:

p₁ = 2(5)

p₁ = +10 kg·m/s

Cart 2:

p₂ = 3(−2)

p₂ = −6 kg·m/s

Total momentum:

p_total = 10 − 6

p_total = +4 kg·m/s

The total momentum is:

4 kg·m/s east

The signs allow opposite directions to be included correctly.

Vector Applications in Engineering

Engineers regularly use vectors to analyse:

• forces on bridges
• forces in cables
• loads on buildings
• aircraft motion
• fluid flow
• vehicle dynamics
• robotic motion
• mechanical systems

For example, several cables may support a load from different directions.

Each cable produces a tension force vector.

The vector sum determines whether the load is:

• in equilibrium
• accelerating
• moving in a particular direction

Vector Applications in Sports

Vectors can describe the motion of:

• footballs
• basketballs
• golf balls
• runners
• swimmers
• cyclists

Suppose a football is kicked at 20 m/s at 35° above the horizontal.

Horizontal velocity:

Vₓ = 20 cos 35°

Vₓ ≈ 16.4 m/s

Vertical velocity:

Vᵧ = 20 sin 35°

Vᵧ ≈ 11.5 m/s

These components can then be analysed separately when studying projectile motion.

Vector Applications in Robotics

A robot may need to move from:

A = (2, 3)

to:

B = (10, 8)

Required displacement:

D = B − A

D = (10 − 2, 8 − 3)

D = (8, 5)

Magnitude:

|D| = √(8² + 5²)

|D| = √89

|D| ≈ 9.43 units

Vector calculations can help the robot determine both the required distance and direction.

Vector Applications in Space Travel

Spacecraft motion depends heavily on vector calculations.

Engineers must consider:

• spacecraft velocity
• gravitational forces
• thrust
• orbital velocity
• changes in direction
• relative motion between planets and spacecraft

A rocket's engines produce thrust vectors.

Changing the direction or magnitude of the thrust changes the spacecraft's velocity.

Because objects in space are constantly moving relative to one another, vector analysis is essential for planning trajectories.

Choosing a Coordinate System

Before solving an applied vector problem, choose positive directions.

A common choice is:

east/right = +x

west/left = −x

north/up = +y

south/down = −y

This makes the signs of the components meaningful.

However, another coordinate system can be used if it makes the problem easier.

The important thing is to remain consistent.

General Strategy for Vector Applications

When solving an authentic vector problem:

Identify the vector quantities.

Determine which quantities have both magnitude and direction.

Choose coordinate directions.

Clearly define positive x and positive y.

Convert vectors into components.

For an angle θ measured from the positive x-axis:

Vₓ = V cos θ

Vᵧ = V sin θ

Add or subtract the components.

Rₓ = sum of x-components

Rᵧ = sum of y-components

Find the resultant magnitude.

|R| = √(Rₓ² + Rᵧ²)

Find the direction.

Use:

tan θ = Rᵧ/Rₓ

Then check the signs of the components to identify the correct direction.

Finally, interpret the result in context.

Include:

• units
• direction
• physical meaning

Worked Authentic Problem: Rescue Boat

A rescue boat can travel at 12 m/s through still water. It heads directly north across a channel while the water current flows east at 5 m/s.

Boat velocity:

V_b = (0, 12) m/s

Current velocity:

V_c = (5, 0) m/s

Ground velocity:

V = V_b + V_c

V = (5, 12) m/s

Magnitude:

|V| = √(5² + 12²)

|V| = √169

|V| = 13 m/s

Direction relative to north:

tan θ = 5/12

θ ≈ 22.6°

Therefore, the boat actually moves:

13 m/s at 22.6° east of north

This is a good example of why direction cannot be ignored.

Worked Authentic Problem: Drone

A drone flies at 15 m/s east relative to the air.

Wind blows at 8 m/s north.

Resultant velocity:

V = (15, 8) m/s

Magnitude:

|V| = √(15² + 8²)

|V| = √289

|V| = 17 m/s

Direction:

tan θ = 8/15

θ ≈ 28.1°

Therefore, the drone moves relative to the ground at:

17 m/s at 28.1° north of east

If the drone needs to travel directly east, its flight controller must compensate for the wind.

Worked Authentic Problem: Towing a Vehicle

Two cables pull a vehicle.

Cable A pulls with 600 N east.

Cable B pulls with 400 N at 60° north of east.

Resolve Cable B:

Bₓ = 400 cos 60°

Bₓ = 200 N

Bᵧ = 400 sin 60°

Bᵧ ≈ 346.4 N

Add components:

Rₓ = 600 + 200

Rₓ = 800 N

Rᵧ = 346.4 N

Magnitude:

|R| = √(800² + 346.4²)

|R| ≈ 872 N

Direction:

tan θ = 346.4/800

θ ≈ 23.4°

Therefore, the combined pulling force is approximately:

872 N at 23.4° north of east

Checking Whether an Answer Makes Sense

Vector calculations should always be checked physically.

For example, if two perpendicular velocities are:

6 m/s

and:

8 m/s

the resultant should be:

• greater than 8 m/s
• less than 14 m/s

The calculated result:

10 m/s

is reasonable.

If your calculation produced 25 m/s, you should immediately suspect an error.

Estimating before calculating is a useful habit.

Common Mistakes

Treating vectors as ordinary numbers

Directions must be considered.

Adding magnitudes instead of components

For vectors acting in different directions, resolve them into components before adding.

Confusing speed and velocity

Speed is scalar. Velocity is vector.

Confusing distance and displacement

Distance describes the total path travelled. Displacement describes the change in position.

Ignoring wind or current

The motion relative to the ground may differ from motion relative to the air or water.

Using sine and cosine incorrectly

Check where the angle is measured from.

Forgetting that bearings are measured from north

Standard mathematical angles are usually measured from the positive x-axis, while bearings are measured clockwise from north.

Ignoring negative components

Negative values usually indicate movement or force in the opposite coordinate direction.

Giving only a magnitude

A complete vector answer often requires both magnitude and direction.

Key Terms

Vector — a quantity with both magnitude and direction.

Scalar — a quantity with magnitude but no direction.

Resultant vector — the single vector representing the combined effect of several vectors.

Component — the part of a vector acting along a particular coordinate direction.

Displacement — the vector change in position.

Velocity — the rate of change of displacement, including direction.

Speed — the scalar rate of distance travelled.

Force — a vector interaction capable of changing an object's motion.

Net force — the vector sum of all forces acting on an object.

Acceleration — the vector rate of change of velocity.

Momentum — the vector quantity p = mv.

Relative velocity — the velocity of one object measured relative to another.

Bearing — a navigation direction measured clockwise from north.

Ground velocity — the actual velocity of an aircraft or other moving object relative to the ground.

Wind correction — adjusting the direction of travel to compensate for wind.

Key Takeaways

• Vectors are essential for modelling quantities that have both magnitude and direction.
• Navigation problems often involve adding displacement or velocity vectors.
• Wind and water currents can change the actual direction and speed of moving objects.
• Forces combine according to vector addition.
• The vector sum of all forces gives the net force.
• If the net force is zero, acceleration is zero.
• Velocity and acceleration are vector quantities.
• Momentum is also a vector because it depends on velocity.
• Angled vectors can be resolved using x = R cos θ and y = R sin θ when θ is measured from the x-axis.
• Resultant magnitude is calculated using |R| = √(Rₓ² + Rᵧ²).
• Navigation bearings require careful attention because they are measured clockwise from north.
• Vector methods are widely used in engineering, aviation, shipping, robotics, sports, and space travel.
• Authentic vector problems should end with an interpretation that includes both magnitude and direction.

Check Your Understanding

1. A person walks 6 km east and then 8 km north. Determine the magnitude and direction of the resultant displacement.
2. A boat moves north at 12 m/s while a current flows east at 5 m/s. Determine the boat's velocity relative to the shore.
3. An aircraft flies east at 300 km/h while wind blows north at 80 km/h. Calculate its ground speed and direction.
4. A swimmer crosses a 90 m wide river at 3 m/s directly across the water. The current flows downstream at 1.5 m/s. Determine:
   • the crossing time
   • the downstream displacement
5. An object experiences forces of 20 N east, 8 N west, 15 N north, and 6 N south. Determine the resultant force.
6. A force of 60 N acts at 40° above the horizontal. Calculate its horizontal and vertical components.

7. A 5 kg object experiences a net force:

   F = (15, 20) N

   Determine its acceleration vector and the magnitude of its acceleration.

8. A 4 kg object moves east at 6 m/s while a 2 kg object moves west at 5 m/s. Determine the total momentum of the two-object system.
9. A drone flies east at 18 m/s while wind blows south at 7 m/s. Determine the drone's ground speed and direction.
10. Two cables pull an object. One produces a force of 500 N east. The other produces a force of 300 N at 45° north of east. Determine the magnitude and direction of the resultant force.
11. A rescue boat can travel at 10 m/s relative to the water. A current of 4 m/s flows east. Explain why the boat cannot simply point directly at a destination directly north if it needs to travel along a straight northward path.
12. Choose a real situation involving navigation, forces, velocity, sports, engineering, or transportation. Identify the vectors involved and explain how vector operations could be used to solve the problem.